Question

im pretty sure ive got this one :) do we need to solve for k first? or does it cancel in the end?

Answer 1

integrate the areas and divide them out .. is my thought

Answer 2

y1 = 1/10 = .1
y2 = 4/10 = .4

A/B

\[A=\int_{0.4}^{1}f(y)~dy\]
\[B=\int_{0.1}^{1}f(y)~dy\]

Answer 3

k(1-y)^4 integrates to -k/5(1-y)^5

which zeros out at 1, so ealuate at -F(a)


-k/5(1-y)^5
----------
-k/5(1-y)^5


(1-.4)^5
-------
(1-.1)^5


(2/3)^5

Answer 4

CORRECT YYOU GET A 0%

Answer 5

the next question is talking about the mean given the PDF

isnt that just:\[\mu=\int x~f(x)\]
but this is going to be fore f(x,y)

Answer 6

this is what i get for the 'graph'

Answer 7

normally i would divide the integral by b-a, but not sure sinc we have a double integration